Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define int long long
using namespace std;
typedef pair<int, int> P;
const int INF = 1e9;
const int V = 100010;

int n, m;
vector<int> G[100010];
int d[100010], dr[100010];

struct dijkstra{
    vector<int> d;
    dijkstra(){}
    dijkstra(int V){
        d.resize(V, INF);
    }
    void calc(int s){
        d[s] = 0;
        priority_queue<P, vector<P>, greater<P> > q;
        q.push(P(d[s], s));
        while(!q.empty()){
            P p = q.top(); q.pop();
            int from = p.second;
            int cost = p.first;
            if(d[from] < cost) continue;
            rep(i, 0, G[from].size()){
                int next = G[from][i];
                int newCost = cost + 1;
                if(d[next] > newCost){
                    d[next] = newCost;
                    q.push(P(newCost, next));
                }
            }
        }
    }
};

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int s, t;
    cin >> n >> m >> s >> t;
    s--; t--;
    rep(i, 0, m){
        int x, y;
        cin >> x >> y;
        x--; y--;
        G[x].push_back(y);
        G[y].push_back(x);
    }
    dijkstra d1(n), d2(n);
    d1.calc(s);
    d2.calc(t);
    int MIN = d1.d[t];
    int ans = 0;
    map<P, int> mp;
    map<int, int> mp1, mp2;
    rep(i, 0, n){
        mp[P(d1.d[i], d2.d[i])]++;
        mp1[d1.d[i]]++;
        mp2[d2.d[i]]++;
    }
    rep(i, 0, MIN){
        // ans += mp[P(i, MIN - i)] * mp[P(i + 2, MIN - i - 2)];
        ans += mp1[i] * mp2[MIN - i - 2];
    }
    cout << ans << endl;
}